# How do you integrate #int xsqrt(1-x^2)dx# from [0,1]?

Substitute:

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If one recognises that the outside of the sqrt is a function of the sqrt differentiated we can proceed by inspection.

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To integrate the function ( \int x \sqrt{1-x^2} , dx ) over the interval ([0,1]), you can use the trigonometric substitution method. Let's substitute ( x = \sin \theta ) and ( dx = \cos \theta , d\theta ). After substitution, the integral becomes ( \int \sin \theta \sqrt{1-\sin^2 \theta} \cos \theta , d\theta ).

Now, ( \sqrt{1-\sin^2 \theta} = \cos \theta ), so the integral simplifies to ( \int \sin^2 \theta , d\theta ) over the interval ([0, \frac{\pi}{2}]).

Using the identity ( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} ), the integral further simplifies to ( \frac{1}{2} \int (1 - \cos 2\theta) , d\theta ).

Integrating term by term, you get ( \frac{1}{2} (\theta - \frac{1}{2} \sin 2\theta) + C ), where ( C ) is the constant of integration.

Now substitute back ( x = \sin \theta ) and ( dx = \cos \theta , d\theta ), and apply the limits ([0,1]). After substituting and simplifying, you'll get the final result for the definite integral over the interval ([0,1]).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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